Question

A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled, (ii) reduced to half?

Hint:

For second-order reactions, the rate is proportional to the square of the concentration. Doubling the concentration increases the rate by a factor of 4, and halving it decreases the rate by a factor of 4.

Answer 1

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Step 1: Understanding the \( S_N2 \) Mechanism.
The \( S_N2 \) mechanism occurs in a single step where the nucleophile simultaneously attacks the electrophilic carbon and displaces the leaving group. This process is highly sensitive to steric hindrance; thus, carbons with fewer substituents are more reactive because they offer easier access to the nucleophile.
2-Bromo-2-methylbutane: Being a tertiary bromide, this compound is highly hindered, making \( S_N2 \) reactions very slow.
2-Bromopentane: As a secondary bromide, it is less hindered than the tertiary one but still more so than a primary bromide, placing it in the middle in terms of reactivity.
1-Bromopentane: This is a primary bromide with minimal steric hindrance, which promotes the fastest \( S_N2 \) reaction.
Thus, the order of increasing reactivity towards \( S_N2 \) displacement is: \[ \text{2-Bromo-2-methylbutane}<\text{2-Bromopentane}<\text{1-Bromopentane}. \]

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